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On the quotient semi-group of a non-commutative semi-group
Murata, Kentaro
Osaka Mathematical Journal. 2(1) P.1-P.5
1950-03
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http://hdl.handle.net/11094/8058
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Osaka University
Osaka Mathematical Journal,
Vol. 2, No. 1, March, 1950
On the Quotient Semi-Group of a Noncommutative
Semi-Group
By Kentaro MURATA
In this short note we remark, following K. Asano'), that a noncommutative semi-group g with a certain condition can be embedded
into the quotient semi-group G. The necessary and sufficient condition
for the existence of the quotient semi-group is the same as the case
of a ring. Moreover if g is a ring, we can define the addition in G
in a natural manner and G is just the quotient ring of g .
Definition 1. An element λ in a semi-group g is called regular, if
the following two conditions are satisfied: 1) αλ —6λ(α, b e g ) implies
α = b and 2) \a — \b(a, b e g ) implies α = b.
If g has the unit, the elements having their inverse elements in g
are obviously regular.
In the following we assume that a semi-group g has regular
elements. It is clear that all regular elements in g form a sub-semigroup β* of g .
Definition 2. Let m be a sub-semi-group of g*. If a semi-group
G which contains g satisfies the next three conditions, we call G a left
quotient semi-group of g by m .
(1) G has a unit 1.
(2) Every element a in m has an inverse or1 in G.
(3) For every x in G, there exists a in m such that ax is contained
in g.
In particular if m = g*, we call G a left quotient semi-group of g .
According to Definition 2, every element s in G is clearly expressible
in the form s = a~la, where α e m and α e g . If g has a left (or right)
unit e, then e — I.3)
Lemma 1. If for every a in g and every a in m there exist af
in m and ar in g such that afa = afa then, for any n elements \ e m
(z —1,..., n) there exist n elements c,: e g (i = l,..., n) satisfying the
folloioing condition:
1) K. Asano, Arithmetische Idealtheorie in nichtkommutativen Rin,gen, Japan. Journ.
Math. 16 (1939) Uber die Quotientenbildung von Schiefringen, Journ. Math. Soc. Japan 1
(1949).
2) e=*el
Kentaro MURATA
Proof. According to the assumption, there exist at in g and a«
in m such that a^ = a2\2. Next, for a2\2 and λ 3 , there exist a2 in
g and α:2 in m such that a± \ = az\2. Next, for α:2 λ2 and λ3 , there
exist a,2 in g and α:3 in m such that α2 c*2λ2 = 0:3X3 e m . By induction
we complete our proof.
Lemma 2. If, under the same assumption as in lemma 1, there
exists one element pair x0, y0 in g such that x0a — y®β£ m(a, βe m)
and Xz a = τ/0 b (a, b 6 g) , then every pair x, y in g satisfying the condition x a = y β G m satisfies xa = yb.
Proof. Putting θ = xQa = yQβ and φ — xa = y β, we take c and
δ in g and m respectively as c 0 = δ 99 e m . Then c 0 = c #0 a = c y0 β
= δφ = Sxa = δyβ, cx0 = 8x, c7/ 0 =δ?/, hence δxa = cxda=^cy0b =
δyb, that is, x a = yb.
Theorem 1. In order that there exists a left quotient semi-group
of g by m, it is necessary and sufficient that for every a e g and every
X e m there exist af G g and λ ' € m satisfying λ'α = α'λ. And such a
left quotient semi-group is uniquely determined by m and g apart from
its isomorphism.
Proof. Let G be a left quotient semi -group of g by m. Then for
a λ~ τ (β e g , λ e m) in G, there exists λ' e m such that λ' a \~l = a' ^ G g ,
namely λ' a = α' λ. Hence the condition is necessary. We show now
the condition is sufficient. First, we assume g has the unit 1. Let G
be the set of all symbols (α, a) , a e m, α e g . We can introduce the
equality of the elements in G as follows: (a, a) is equal to (β, 6) if
and only if xa = yb for every x and y satisfying xa = y βζ m, #, ye
g . Then, according to Lemma 2, in order that (α, α) = (/8, 6) it is
sufficient that there exists at least one element pair #, ?/ in g satisfying xa = yb and xa = yβem. As we can then readily prove, the
above-defined equality fulfils the equivalence relation. In particular
(λα, λα) = (tf, α)(λeιn). Now we define the multiplication of the
elements in G as follows:
(α, α)(/9, δ) = (/3'α, α'6), β'a = a'β, /3'6m, α' 6 g .
The product is independent of the choice of α' e g and /β' e m . For if
β»a — a'f β, β" em, a" e g , then taking u and δ such that u βf = δ β" e
m ( w G g , S e m ) , we get uβfa= δβ"aem, and
Hence ua' = δa", uafb=-δa»b, that is, (/3'α:, α'6) = (/S//α, α^δ) . In
Quotient Semi-Groub of a Noncomiitathe Semi-Group.
3
particular, if the product of a and β is commutative, then (α, a) (β, 6)
= (/3α, α6). Further, if (a, a) — (a,, αO and 03, 6) — (β , &0 then
f
(<*, α)(/3, 6) = (αj, αJCA, 6j). For if we choose c, ^, a' in g and β
in m satisfying y = cβ = c1 β, em, /3' α = α'7, then (or, α)(/3, &) = (/3'α,
α' c b) = 09' α, α' c, 6,) = (α, α) (A , & A ). Similarly (α, α) (β,, &0 = (<*ι , α,)
(A , &i). One can easily verify the associative law of the multiplication introduced above. Hence G is a semi-group. The mapping α->
(1, a) gives an isomorphism of g into G. Identifying a and (1, a), we
can see that G contains g , 1 = (I, 1) is the k unit of G, every element
a = (1, a) in m has an inverse (or, 1) in G, and
a (a, a) = (1, α) (α, α) = (α, a a) — (1, α) = α .
Hence G is a left quotient semi-group of g by m. Secondly, if g has
no unit, then we add it to g , and denote the new by g'. Then there
exists a left quotient semi-group G of g' by m. And it is easily seen
that G is a left quotient semi-group of g by m. Finally let G' be an
arbitrary left quotient semi-group of g by m, then every element in
G' is expressible as the form a~λ a where a is in m and α in g . The
mapping αr1^—>(α:, a) gives an isomorphism of Gf onto G, q. e. d.
Corollary 1. In order that there exists a left quotient semi-group
of a semi-group g , it is necessary and sufficient that for every regular
element λ and every element a in g there exist a' in g and a regular
element λ' satisfying af\ — \fa.
Corollary 2. Let all elements in a semi-group g be regular. In
order that there exists a left quotient group of g , it is necessary and
sufficient that for any two elements a, β in g there exist another two
elements af, βf in g satisfying afa = β'β.
Lemma 3. Let G be a left quotient semi-group of g by m. Let
further x,i:(i = ί, ..., n) be any n elements in g , then we can choose
an element γon such that 7 # t € g ( i = l, • • • , n').
Proof. There exist ab in m such that at x{ € g (i = 1, ... , n). And
if we take, according to Lemma 1, γ = CLCCL ==... = cnanem, then γxt
6 f l ( i = l, ..., n).
Now, we shall consider a quotient system of some algebraic system.
Let o bs a semi-group with regular elements, and m a sub-semi-group
of o consisting of regular elements. Let further o has . besides the
multiplication, binary operations denoted by o , which satisfy the
following conditions:
[1] If a and 6 (α, b e o) are composable, (with respect to o) then
ca and cb are composable, also ac and be are composable. And ca o cb
— c (a o ί>), ac o be = (a o &) c .
4
Kentarό MURATA
[2] If, for some 7 e m, 7 α and 7 b (a, b € o) are composable, then c&
and b are composable, also if ay and 6,7 are composable then so are
a and 6.
For example, let o be a (noncommutative or commutative) ring,
and o the addition, then the above-mentioned conditions are of course
fulfilled.
Theorem 2. Let S be a left quotient semi-group of o by m. //
there exists an element X E m , such that \x and \y (x, 2/GS) are contained in o and composable, we define the composition of x and y by
X o 2/ = X ~ l (\X o X?/) .
(*)
Then S becomes an algebraic system of the same kind of o .
Proof. If x and y (where x, y e S) are composable in the sense of
the above-defined (*) , then x o y is uniquely determined independently
of the choice of X in m. For if μ x and μ y (where μxζo , μyeo,
/A em) are composable, then, by taking η and c satisfying 'η μ — c X =
7 e m (η e m, c e o ) , we get
7 X - 1 (λ X o λ ?/) = C X λ""1 (λ # o X T/) =c (X x o X 2/)
ημy=η(μXoμy')
= y μ~l (μX & μy} .
Namely X" 1 (λ x o x ?/) = μ-1 (μx* μy}.
Next, if a; and # in S are composoble, then so are zx and ^ (»
and . « α? o ^ ?/ = z (x ° 2/) . By the hypothesis, there exists λ in m
such that x oy — \~l (\χ o\yy. If we choose a, τ and a such that
azeo , a - z # e o , α s y e o (α e m) and r (a z) = a λ (r e m, α e o) , then
σ z x = a\x, σ z y = a\y (σ = τ a) . Since a λ x = σ ^ α; and a λ £/ =
cr 2 ?/ are composable in o , z x and 2; y are composable in S, and
l
z (x ° y} = z \'λ (λ x o λ 2/) =' σ~ α (λ α; o x ?/)
= σ" 1 (α X x o c& X T/) = σ~l ( σ z χ o σ z y ) = zχozy,
(σ e m) .
If # and y in S are composable, then x z and T/ « are composable,
and (x o y^z = x z ° y z . By the hypothesis, there exists X e m such that
x o y r=z \-ι (x x o x T/) . If we take μ and α: such that μ z e o (μ e m) ,
α X # μrl e o and <# X ?/ μ~l € o (α e m), then x o y = σ~λ (σ xoσy), σ = a\.
According to the latter part of the condition [2], σx μ~l and σy μ~l
are composable in o . Hence
(X o T/) z = or~ l (σ X o σy) Z = σ" 1 (cr a; y^" 1 μ o <r y μ~l μ} Z
l
τ
= σ~l (σ 0? μΓ oσ 2/ /^~ ) μZ — σ~l (σ X μ~l μ Z o cr y μ~l μ Z)
= cr"1 (cr XZ oσ y z) =^X Z o y z.
Quotient Semi-Group of a Noncommutaίive Semi-Group.
5
Every element in m has an inverse in S. Then condition [2] is
therefore easily obtained from the condition [1]. Thus our theorem
is proved.
Let o be a noncommutative ring containing non-nilfactors. The
non-nilfactor is clearly a regular element in o in the sense of Definition 1. Hence we have
Corollary**) Let m be a sub-semi-group consisting of non-nilfactors
in a given ring o . In order that there exists a left quotient ring S
of o by m, it is necessary and sufficient that, for every a 6 m and every
a G o , there exist a' e o and af e m satisfying af a = a! a.
Putting the word right in place of left in the above-mentioned
argument, we can argue similarly as above. But the existence of a
left quotient semi-group and a right one are independent, and if there
exist both, then they are the same4).
(Received June 8, 1949)
<*), *)
K. Asano, 1. c.